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Ket [755]
3 years ago
9

Vector has a magnitude of 4.40 m and is directed east. Vector has a magnitude of 3.40 m and is directed 39.0° west of north. Wha

t are (a) the magnitude and (b) the direction (counterclockwise from east) of ? What are (c) the magnitude and (d) the direction (counterclockwise from east) of ?
Physics
1 answer:
Olenka [21]3 years ago
8 0

Answer:

Magnitude = 4.056 m

Direction = 42.3⁰

Explanation:

The vector is resolved in terms of the vertical and horizontal components. Let's look each of these separately.

The vector 4.40 is directed East. This automatically becomes a horizontal component.

But we know that there is a vector 3.40 North West. The angle the vector makes with the horizontal is 61⁰.

Resolving the vectors should yield the horizontal and vertical components:

Horizontal components

The first component is 4.40 m

The second one is derived by resolving 3.40 to the horizontal like this 3.40 × - cos 61⁰ = -1.648 m

Adding the horizontal component gives 4.40 m + ( -1.648 m) = 2.752 m

Vertical components

Resolve 3.40 with the angle 61⁰ like this: vertical comp = 3.41 × sin 61

                                                                                           = 2.98 m

The magnitude is given by √[(2.98)²+ (2.752)²] = 4.056 m Ans

The direction us given by tan⁻¹ (2.98/2.752) = 42.3⁰ Ans

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The engine in an imaginary sports car can provide constant power to the wheels over a range of speeds from 0 to 70 miles per hou
puteri [66]

Answer:

t=5.3687\ s  is the time taken by the car to accelerate the desired range of the speed from zero at full power.

Explanation:

Given:

Range of speed during which constant power is supplied to the wheels by the car is 0\ mph\ to\ 70\ mph.

  • Initial velocity of the car, v_i=0\ mph
  • final velocity of the car during the test, v_f=32\ mph=14.3052\ m.s^{-1}
  • Time taken to accelerate form zero to 32 mph at full power, t=1.2\ s
  • initial velocity of the car, u_i=0\ mph
  • final desired velocity of the car, u_f=64\ mph=28.6105\ m.s^{-1}

Now the acceleration of the car:

a=\frac{v_f-v_i}{t}

a=\frac{14.3052-0}{1.2}

a=11.921\ m.s^{-1}

Now using the equation of motion:

u_f=u_i+a.t

64=0+11.921\times t

t=5.3687\ s is the time taken by the car to accelerate the desired range of the speed from zero at full power.

8 0
3 years ago
A car turns a certain curve of radius 24.98 m with constant linear speed of
Anastaziya [24]

Answer:

3525.19 kg

Explanation:

The computation of the mass of the car is shown below:

As we know that

Fc = m × V^2 ÷ R

m = Fc × R ÷ V^2

Provided that:

Fc = 34.652 kN = 34652 N

R = Radius = 24.98 m

V = speed = 15.67 m/s

So,

m = 34652 × 24.98 ÷ 15.67^2

 = 3525.19 kg

7 0
3 years ago
How much force is needed to accelerate a bicycle and rider with a total mass of 50 kg at a rate of 2.0 m/s2 ?
True [87]

Answer:

The answer is D 100 newton

Explanation:

2.0m/s2 is d acceleration while the 50kg is the mass. Force = mass x acceleration. So f=50x2.so force is 100 newton

8 0
3 years ago
When the rubber sheet is pulled down, the volume of the bell jar is ____________ and the pressure in the bell jar is ___________
Makovka662 [10]

Answer:

increases, decreases, respiration

8 0
2 years ago
1. What could be the energy of a flying aeroplane?<br><br> 2. Is work the same as power?
Doss [256]

Explanation:

The total energy of an aircraft flying in the atmosphere can be calculated using equation 1. [2]

E = ½ m v2 + mgh

A Boeing 737-300 has a maximum takeoff weight of 5.65 × 104 kg, a cruise altitude of h = 10,195 m, and cruise speed of 221 m/sec. Inserting these numbers into the above equation, we obtain 7.03 GJ for the energy at cruise conditions. [3] However, the engines mounted onto the wings of the plane are required to provide additional energy per time, power, in order to keep the aircraft flying at a constant altitude and speed

Work is the energy needed to apply a force to move an object a particular distance, where force is parallel to the displacement. Power is the rate at which that work is done.

7 0
3 years ago
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